heureka

Find the value of
$S_n=\dfrac{1}{2} + \dfrac{3}{2^2} + \dfrac{5}{2^3} + ... + \dfrac{2n-3}{2^{n-1}} + \dfrac{2n-1}{2^n}$

$\text{Let $x=\dfrac12$}$

$\begin{array}{|rcll|} \hline S_n&=&x + &3x^2 + 5x^3+ 7x^4 + ... + (2n-3)x^{n-1} + (2n-1)x^n \\ xS_n&=& &1x^2 + 3x^3+ 5x^4 + ... + (2n-5)x^{n-1} + (2n-3)x^n+(2n-1)x^{n+1} \\ \hline x-xS_n &=& x + &2x^2 + 2x^3+ 2x^4 + ... + 2x^{n-1} + 2x^n -(2n-1)x^{n+1} \\ S_n(1-x) &=& x + &2*(x^2 + x^3+ x^4 + ... + x^{n-1} + x^n) -(2n-1)x^{n+1} \\ &&& \boxed{x^2 + x^3+ x^4 + ... + x^{n-1} + x^n = \dfrac{x^2-x^{n+1} } {1-x} } \\ S_n(1-x) &=& x + &2*\dfrac{x^2-x^{n+1} } {1-x} -(2n-1)x^{n+1} \quad | \quad 1-x=1-\dfrac12=\dfrac12 \\ S_n*\dfrac12 &=& x + &4(x^2-x^{n+1}) -(2n-1)x^{n+1} \quad | \quad * 2 \\ S_n &=& 2x + &8(x^2-x^{n+1}) -2(2n-1)x^{n+1} \\ S_n &=& 2x + &8x^2-8x^{n+1} -2(2n-1)x^{n+1} \\ S_n &=& 2x + &8x^2-2x^{n+1}\Big(4 +(2n-1)\Big) \\ S_n &=& 2x + &8x^2-2x^{n+1}(2n+3) \\ S_n &=& 2x \Big(& 1 + 4x-x^n(2n+3)\Big) \quad | \quad x=\dfrac12 \\ S_n &=& &1 + 2-\dfrac{2n+3}{2^n} \\ \mathbf{S_n} &=& \mathbf{3-}&\mathbf{\dfrac{2n+3}{2^n}} \\ \hline \end{array}$

A sequence is defined by 

$\begin{array}{|rcll|} \hline t_{345} &=& 2 &\quad 345 = 4*86+1 \\ \hline \end{array}$

Find the general term of the following sequences:

a)$7, 13, 19, 25, 31, \ldots$
$a_n=6n+1$

b) $4, –6, 9, -\frac{27}{2}, \ldots$
$a_n=4*(-\frac{3}{2})^{n-1} \\ a_n=a_{n-1}*(-\frac{3}{2})$

Compute the value of
$1 - 2 + 3 - 4 + \dots + 2019 - 2020 + 2021$.

$\begin{array}{|rcll|} \hline s &=& 1 - 2 + 3 - 4 + \dots + 2019 - 2020 + 2021 \\ s&=& 2021 - 2020 +2019 - \dots -4+3-2+1 \\ \hline 2s &=& 2022-2022+2022-2022 + \dots + 2022-2022+2022 \\ 2s &=& 0+0+ \dots +0+2022 \\ 2s &=& 2022 \\ s &=& 1011 \\ \hline \end{array}$

$1 - 2 + 3 - 4 + \dots + 2019 - 2020 + 2021 = \mathbf{1011}$

Find the value of series
$1 + 4 + 2 + 8 + 3 + 12 + 4 + 16 + \cdots + 24 + 96 + 25 + 100.$

$\begin{array}{|rcll|} \hline 1+4 &=& 5 \\ 2+8 &=& 10 \\ 3+12 &=& 15 \\ 4+16 &=& 20 \\ \ldots \\ 24 + 96 &=& 120 \\ 25 + 100 &=& 125 \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline && 1 + 4 + 2 + 8 + 3 + 12 + 4 + 16 + \cdots + 24 + 96 + 25 + 100\\ &=& 5+10+15+20 + \cdots +120+125 \\ &=& 5*(1+2+3+4+\cdots +24+25) \\ && \boxed{ 1+2+3+4+\cdots +24+25 = \frac{1+25}{2}*25 \\ =13*25 \\ = 325 } \\ &=& 5*325 \\ &=& \mathbf{1625} \\ \hline \end{array}$

Let $a,~b,~c$ be the roots of $x^3 - x + 4 = 0$.
Compute $(a^2 - 1)(b^2 - 1)(c^2 - 1)$.

Let $a,~b,~c$ be the roots of $x^3 - x + 4 = 0$.
Compute $(a^2 - 1)(b^2 - 1)(c^2 - 1)$.

By Vieta:

$\begin{array}{|rccc|} \hline x^3 &\underbrace{+0*x^2}_{0=-(a+b+c)} &\underbrace{-1*x}_{-1=ab+ac+bc} & \underbrace{+4}_{4=-abc} \\\\ \hline \end{array} \begin{array}{|rcll|} \hline abc &=& -4 \\ ab+ac+bc&=& -1 \\ a+b+c &=& 0 \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline a+b+c &=& 0 \\ (a+b+c)^2 &=& 0^2 \\ a^2+b^2+c^2+2*(ab+ac+bc) &=& 0 \\ a^2+b^2+c^2+2*(-1) &=& 0 \\ \mathbf{a^2+b^2+c^2} &=& \mathbf{2} \\ \hline \end{array} \\ \begin{array}{|rcll|} \hline ab+ac+bc &=& -1 \\ (ab+ac+bc)^2 &=& (-1)^2 \\ a^2b^2+a^2c^2+b^2c^2+2*(a^2bc+ab^2c+abc^2) &=& 1 \\ a^2b^2+a^2c^2+b^2c^2+2*abc*(a+b+c) &=& 1 \\ a^2b^2+a^2c^2+b^2c^2+2*abc*(0) &=& 1 \\ \mathbf{a^2b^2+a^2c^2+b^2c^2} &=& \mathbf{1} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline && \mathbf{(a^2 - 1)(b^2 - 1)(c^2 - 1)} \\\\ &=& a^2b^2c^2-(a^2b^2+a^2c^2+b^2c^2)+a^2+b^2+c^2-1 \\ &=& (-4)^2-(1)+2-1 \\ &=& 16-2+2 \\ &=& \mathbf{16} \\ \hline \end{array}$

$(a^2 - 1)(b^2 - 1)(c^2 - 1) = \mathbf{16}$

Let $a$, $b$, and $c$, be nonzero real numbers such that $a+b+c=0$.
Compute the value of
$a\left(\dfrac 1b + \dfrac1c\right) + b\left(\dfrac1a + \dfrac1c\right) + c\left(\dfrac1a+\dfrac1b\right)$.

$\begin{array}{|rcll|} \hline a+b+c &=& 0 \\ a+b &=& -c \qquad (1) \\ a+c &=& -b \qquad (2) \\ b+c &=& -a \qquad (3) \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline && \mathbf{a\left(\dfrac 1b + \dfrac1c\right) + b\left(\dfrac1a + \dfrac1c\right) + c\left(\dfrac1a+\dfrac1b\right)} \\\\ &=& \dfrac ab + \dfrac ac + \dfrac ba + \dfrac bc + \dfrac ca+\dfrac cb \\\\ &=& \dfrac ba + \dfrac ca + \dfrac ab + \dfrac cb + \dfrac ac+\dfrac bc \\\\ &=& \dfrac {b+c}a + \dfrac {a+c}b + \dfrac {a+b}c \\\\ &=& \dfrac {-a}a + \dfrac {-b}b + \dfrac {-c}c \\\\ &=& -1-1-1 \\\\ &=& \mathbf{-3} \\ \hline \end{array}$

$a\left(\dfrac 1b + \dfrac1c\right) + b\left(\dfrac1a + \dfrac1c\right) + c\left(\dfrac1a+\dfrac1b\right)=-3$

What is the value of $(1 + i)^5 - (1 - i)^5$?

$\begin{array}{|rcll|} \hline (1+i)^5-(1-i)^5 &=& \left((1+i)^2\right)^2(1+i)-\left((1-i)^2\right)^2(i-1) \\ &=& \left(1+2i+i^2\right)^2(1+i)-\left(1-2i+i^2\right)^2(1-i) \\ &=& \left(1+2i-1\right)^2(1+i)-\left(1-2i-1\right)^2(1-i) \\ &=& \left(2i\right)^2(1+i)-\left(-2i\right)^2(1-i) \\ &=& 4i^2(1+i)-4i^2(1-i) \quad | \quad i^2 =-1 \\ &=& -4(1+i)+4(1-i) \\ &=& -4-4i+4-4i \\ \mathbf{(1+i)^5-(1-i)^5} &=& \mathbf{-8i} \\ \hline \end{array}$

Hello jleung,